Question

The equation for the reaction between zinc and dilute hydrochloric acid is Zn 2HCi ZnCi2 + H2
What the volume of hydrogen gas produced by 3.25g of zinc?

Answer 1

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.

Explanation:

Given: The equation for the reaction between zinc and dilute hydrochloric acid is:

Zn + 2HCl → ZnCl2 + H2

Find: Volume of hydrogen gas produced by 3.25g of zinc.

Solution:  

Molar mass of Zn = 65.4 g/mol

So 65.4 g of zinc liberates 1 mol of hydrogen gas (22.4 L) at STP

Therefore 3.25 g of zinc will liberate:  

22.5 / 65.4 * 3.5 = 1.20 L of hydrogen gas at STP

Answer 2

To find:

• Amount of hydrogen gas when 3.25 g of zinc is used in the equation?

Calculation:

$\boxed{\bf Zn+2HCl\rightarrow ZnCl_{2}+H_{2}}$

• Now, in this question, we will consider Zn as the limiting reagent.

• Also, we will simply use UNITARY METHOD in this kind of stoichiometric questions.

So, 1 mole of Zn (65 grams) produces 22.4 L of hydrogen gas at STP.

Now, 3.25 grams of Zn will produce $\dfrac{22.4}{65}\times 3.25 = 1.12\:Litres$

So, volume of Hydrogen gas produced is 1.12 Litres.